12x^2-38x+25=0

Solution for 12x^2-38x+25=0 equation:

12x^2-38x+25=0

a = 12; b = -38; c = +25;
Δ = b2-4ac
Δ = -382-4·12·25
Δ = 244
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{61}}{2*12}=\frac{38-2\sqrt{61}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{61}}{2*12}=\frac{38+2\sqrt{61}}{24}
$